高中数学 342(对数)课件 北师大版必修1 课件VIP免费

25/2/25对数25/2/251.进一步熟悉对数的运算性质.2.熟练运用对数运算性质.3.掌握化简、求值技巧.4.培养学生数学应用意识.教学目标25/2/251.基本性质:若a>0且a≠１,N>0,则2.常用性质:若a>0且a≠１,则(1)loga1=0(2)logaa=13.运算性质:若a>0且a≠１,M>0,N>0,则baNabaNalog21logMnMNMNMNMMNanaaaaaaaloglog3logloglog2logloglog125/2/25例1.求下列各式的值(1)log0.41(2)log2(47×25)(3)lg510001log14.0：解195722log2log2log4log24log25272252725725210lg5210lg51100lg32525/2/2532log2;log1log,log,log.4zyxzxy：zyxaaaaa表示下列各式用例zyxzyxzyxzyxzyxzxyzxyaaaaaaaaaaaaaaalog31log21log2loglogloglogloglog2loglogloglogloglog1:323232解25/2/252.1lg10lg38lg27lg3;9lg243lg2;18lg7lg37lg214lg1:.5计算例01lg1837714lg18lg7lg37lg14lg18lg7lg37lg214lg02lg3lg27lg3lg27lg27lg2lg)23lg(7lg)3lg7(lg272lg18lg7lg37lg214lg1222、、解法二解法一25/2/252312lg23lg12lg23lg231023lg10lg32lg3lg2.1lg10lg38lg27lg3253lg23lg53lg3lg9lg243lg2221321325解25/2/25的值求已知例44.1lg.4771.03lg,3010.02lg:61582.01310.024771.0212lg23lg21023lg44.1lg:212解25/2/25.,loglog:7xbcxaa求已知例bbcbcacaaaxaaloglog::由对数定义可知解法一bbaaaacxacxbcxbcx,logloglog由对数定义知即解法二由已知移项可得bbabaaabaacxacacxablogloglogloglog:解法三25/2/25课时小结通过本节学习，大家应熟悉对数的运算性质应用；并掌握一定的解题技巧，积累一定的解题经验。25/2/25